tag:blogger.com,1999:blog-8231784566931768362.post1316133025844739615..comments2023-09-09T08:21:55.454-04:00Comments on MathNotations: Last minute PSAT Prep... A Challenge for the rest of us!Dave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-8231784566931768362.post-40602368849274334712007-10-16T21:11:00.000-04:002007-10-16T21:11:00.000-04:00Before 6am, after 6am, I've stopped tracking when ...Before 6am, after 6am, I've stopped tracking <B>when</B> they occur. All I know is the number of occurrences is in one-to-one correspondence with the reals.<BR/><BR/><I>That's about the extent of the math I'll be able to add to the conversation at this point.</I>Jackie Ballarinihttps://www.blogger.com/profile/16859831037023994373noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-9373911154828736742007-10-15T16:54:00.000-04:002007-10-15T16:54:00.000-04:00Jackie,That's weird. All of my mistakes occur AFTE...Jackie,<BR/>That's weird. All of my mistakes occur AFTER 6 AM!<BR/>DaveDave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-37849212819616833832007-10-15T16:21:00.000-04:002007-10-15T16:21:00.000-04:00argh, that's what I get for trying to comment befo...argh, that's what I get for trying to comment before 6:00 in the morning. I meant, they have trouble finding f(3), when f(x) = .....<BR/><BR/>Sorry.Jackie Ballarinihttps://www.blogger.com/profile/16859831037023994373noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-11232511157100514802007-10-15T12:11:00.000-04:002007-10-15T12:11:00.000-04:00Hey tc!I knew you wouldn't miss that opportunity! ...Hey tc!<BR/>I knew you wouldn't miss that opportunity! What's really interesting is that many students appreciate the pattern of triangular numbers for these types.<BR/>For sum of 3:<BR/>300; 210,201; 120,102,111<BR/>1+2+3 = 6 = 4C2 = (3)(2)/2<BR/>In general, the nth triangular number is nC2 = n(n-1)/2.<BR/>Works for a sum of 6 as well, of course.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-36734178094400505422007-10-15T09:42:00.000-04:002007-10-15T09:42:00.000-04:00History repeats itself as to the methods that can ...History repeats itself as to the methods that can be used. Thus, the answers to 2(a) and 2(b) are 4C2 and 7C2.<BR/><BR/>TCUnknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-15251897732736945492007-10-15T09:07:00.000-04:002007-10-15T09:07:00.000-04:00Wait a minute: what does f(3)=2x+4 mean? Are you ...Wait a minute: what <I>does</I> f(3)=2x+4 mean? Are you using x as a constant there?mathmomhttps://www.blogger.com/profile/05869925405540832241noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-11506224346043949652007-10-15T06:41:00.000-04:002007-10-15T06:41:00.000-04:00I'm going to back Dave up on this one (not that he...I'm going to back Dave up on this one (not that he needs it). Function notation is confusing for kids.<BR/><BR/><I>f(3)</I>=2x+4<BR/><BR/>Many (of my) students <B>still</B> think this means 3 = 2x +4.Jackie Ballarinihttps://www.blogger.com/profile/16859831037023994373noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-80139160475120654902007-10-15T00:11:00.000-04:002007-10-15T00:11:00.000-04:00Expressing the problem in function notation immedi...<I>Expressing the problem in function notation immediately raises the difficulty level since many high school students are still intimidated by this, even Algebra 2 students (experience talking).</I><BR/><BR/>Really? I would have guessed that having a real number in the problem (instead of k, as in part a) would make it easier, and the function notation is so peripheral there. But I don't have experience with Algebra II kids, so I'll take your word for it. :)mathmomhttps://www.blogger.com/profile/05869925405540832241noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-3909862281004285322007-10-14T18:53:00.000-04:002007-10-14T18:53:00.000-04:00Mathmom--You rarely if ever miss something! I agre...Mathmom--<BR/>You rarely if ever miss something! I agree that this is straightforward, however there are several points to be made here:<BR/>(1) Expressing the problem in function notation immediately raises the difficulty level since many high school students are still intimidated by this, even Algebra 2 students (experience talking). The function piece is what makes this unlikely on the PSAT.<BR/>(2) I really had an ulterior motive for selecting this problem. It encourages the use of multiple representations (aka, 'Rule of Four') by the instructor:<BR/>(a) Function table approach (with or without the graphing calculator)<BR/>(b) Verbal interpretation, i.e., x cannot be negative using your explanation)<BR/>(c) Graphical/visual approach: The parabola is only below the x-axis for x-values between the intercepts<BR/>(d) Algebraic solution using 'sign-charts' (or the myriad of names for this):<BR/>+++++++++|--------|++++++++<BR/> 0 2007<BR/>I agree with all of your comments except for the fact that, in actual practice, student performance on these kinds of questions is surprisingly low. That's why it typically appears among the last 5 questions of a section on the exam. Remember, the Educational Testing Service has a pretty good idea of levels of difficulty (and therefore problem placement) even before they administer the real test! This is because these questions have been placed in an experimental section prior to the actual 'live' implementation (the 'operational' version). If, say, 75% of the students answered this question incorrectly on an experimental section, ETS places this question among the last 5 on the actual test. In some cases, questions are rejected based on the field-test statistics (reliability, p-value too low, etc.).Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-76019459068772705802007-10-14T18:02:00.000-04:002007-10-14T18:02:00.000-04:00Unless I'm missing something, 4b seems like one of...Unless I'm missing something, 4b seems like one of the easier problems to me. if we want x^2 - 2007x < 0, first we note that x must be positive (since if x were negative, then x^2 would be positive and -2007*x would also be positive). So, we need 2007x > x^2 or 2007 > x (we can divide through by x without worrying because we know it's positive). Checking x=0, we see that f(x) = 0. So 0<x<2007 which range contains 2006 positive integers. Even kids not used to working with inequalities, this seems pretty easy to "eyeball" to me. you want 2007*x to be bigger than x*x which will be true as long as 2007 is bigger than x.mathmomhttps://www.blogger.com/profile/05869925405540832241noreply@blogger.com